Expanding the Integration problem.
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How does this work like when you expand the integration how did it result in -1/4(b+a)^2?
If someone could help explain this part to me that would be greatly appreciated, thanks a bunch.
1 answer
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This is just middle school algebra.
$\frac{1}{b-a} \left[ -\dfrac{1}{2}(a + b){\color{red}{(b^2 - a^2)}} + \dfrac{1}{4}(b+ a)^2(b-a)\right] \equiv \frac{1}{b-a} \left[ -\dfrac{1}{2}(a + b){\color{red}{(a + b)(b - a)}} + \dfrac{1}{4}(b+ a)^2(b-a)\right] \equiv \frac{1}{b-a} \left[ -\dfrac{1}{2}{\color{limegreen}{(a + b)^2(b - a)}} + \dfrac{1}{4}{\color{limegreen}{(a + b)^2(b - a)}} \right]$
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